The range of 'y, such that the equation in x, `y + cos x = sin x` has a real solution is

Options

A. `[-2, 2]`
B. `[- sqrt(2), sqrt(2)]`
C. `[-1, 1]`
D. `[-1//2, 1//2]`

Explanation

To solve the equation \( y + \cos x = \sin x \) for the range of \( y \) such that there is a real solution in \( x \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y + \cos x = \sin x \]
Rearranging this gives:
\[ y = \sin x - \cos x \]

### Step 2: Expressing in a Different Form
Next, we can express \( \sin x - \cos x \) in a more manageable form. We can factor out \( \sqrt{2} \) by multiplying and dividing by \( \sqrt{2} \):
\[ y = \sin x - \cos x = \frac{1}{\sqrt{2}}(\sqrt{2} \sin x - \sqrt{2} \cos x) \]

### Step 3: Using Trigonometric Identities
We recognize that \( \sqrt{2} \sin x - \sqrt{2} \cos x \) can be rewritten using the sine subtraction formula. We know that:
\[ \sqrt{2} \sin x - \sqrt{2} \cos x = \sqrt{2} \left( \sin x - \frac{1}{\sqrt{2}} \cos x \right) \]
This can be expressed as:
\[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \]
Thus, we have:
\[ y = \frac{1}{\sqrt{2}} \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) = \sin \left( x - \frac{\pi}{4} \right) \]

### Step 4: Finding the Range of \( y \)
The sine function oscillates between -1 and 1. Therefore, the range of \( y \) is:
\[ -1 \leq y \leq 1 \]

### Step 5: Relating to the Original Equation
We need to find the range of \( y \) such that the equation has a real solution. Since we have expressed \( y \) in terms of \( \sin \left( x - \frac{\pi}{4} \right) \), we can conclude that \( y \) must lie within the bounds of the sine function.

### Final Result
Thus, the range of \( y \) such that the equation \( y + \cos x = \sin x \) has a real solution is:
\[ [-\sqrt{2}, \sqrt{2}] \]